Questions 3 through 8: Given names, formulas are to be written; given formulas, names are to be written. Germanium (Z = 32), fluorine, argon, gallium (Z = 31) Concept introduction: Mainly, the elements are found as single atom particles in the universe. For example, the element which exists in the gas phase as single atoms at temperatures is Helium and the element which found in the solid phase is Lithium. Predominantly, the elemental symbol is used to denote this one atom and further explain that these atoms are uncombined in nature.

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Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100

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Chapter 6, Problem 8E

Interpretation Introduction

Interpretation:

Questions 3 through 8: Given names, formulas are to be written; given formulas, names are to be written.

Germanium (Z = 32), fluorine, argon, gallium (Z = 31)

Concept introduction:

Mainly, the elements are found as single atom particles in the universe. For example, the element which exists in the gas phase as single atoms at temperatures is Helium and the element which found in the solid phase is Lithium. Predominantly, the elemental symbol is used to denote this one atom and further explain that these atoms are uncombined in nature.

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Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100

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